Answer
$x=2+\sqrt{5}i$ or $x=2-\sqrt{5}i$
Work Step by Step
$ x(x-3)=x-9\qquad$... use the distributive property: $a(b+c)=ab+ac$.
$ x^{2}-3x=x-9\qquad$...add $(-x+9)$ to both sides.
$ x^{2}-3x-x+9=x-9-x+9\qquad$...add like terms.
$ x^{2}-4x+9=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-4,\ c=9$
$ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-4,\ a$ for $1$ and $c$ for $9$.
$ x=\displaystyle \frac{-(-4)\pm\sqrt{(-4)^{2}-4\cdot(9)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$x=\displaystyle \frac{4\pm\sqrt{16-36}}{2}$
$x=\displaystyle \frac{4\pm\sqrt{-20}}{2}$
$ x=\displaystyle \frac{4\pm\sqrt{-1\cdot 4\cdot 5}}{2}\qquad$... write in terms of $i$. ($\sqrt{-1}=i$)
$x=\displaystyle \frac{4\pm 2i\sqrt{5}}{2}$
$ x=2\pm\sqrt{5}i\qquad$... the symbol $\pm$ indicates two solutions.
$x=2+\sqrt{5}i$ or $x=2-\sqrt{5}i$
